When the drain voltage exceeds the pinch-off voltage, the effective channel length of the transistor is reduced. In other words, the layer of mobile inversion charge (the electrons in an N MOSFET) is pushed away from the drain. So the question is: why does current still flow if the layer of inversion charge does not reach the drain?Group of answer choicesCurrent does not flow - the transistor is off in pinch-off.The current is actually carried by mobile holes in an N MOSFET to the drain.The electrons diffuse to the drain when they reach the depletion region formed by the drain-substrate PN junction. The electric field in the reverse-biased PN junction formed by the drain and substrate sweeps the mobile charge (electrons in an N MOSFET) to the drain when the mobile charge (electrons in an N MOSFET) reach the depletion region of the drain-substrate PN junction.

Arrange the six figures labelled (A) to (F) in the proper sequential order to show: (1) PN junction depletion region around the drain and source(2) Small positive voltage applied to gate creating a depletion region beneath the gate by pushing away the mobile holes.(3) The gate voltage is at the threshold voltage where the number of mobile electrons and holes beneath the gate are approximately equal.(4) The gate voltage is above the threshold voltage and there are electrons beneath gate forming a "inversion" layer of charge.(5) The drain voltage is increased to the Pinch-off voltage where the reverse-biased depletion region between the n-type drain and p-type substrate increases so that the inversion layer of charge disappears (or is pinched-off) at the drain.(6) The drain voltage is increased beyond the Pinch-off voltage and the effective channel length beneath the gate decreases. (A) (B)(C)  (D) (E) (F)Group of answer choicesB,D,A,F,C,EE,C,F,A,B,DB,D,A,F,E,CD,B,F,A,C,E

In a reverse-biased P-N junction diode, the current is primarily due to:Options :Electron flow from N to P regionHole flow from P to N regionMajority charge carriersMinority charge carriers

8.2.1 The Basic MOSFET Current SourceFigure 8.1 shows the circuit of a simple MOS constant-current source. The heart of the circuitis transistor Q1 , the drain of which is shorted to its gate, 1 thereby forcing it to operate in thesaturation mode withID1 = 12 k′n( WL)1(V GS − Vtn)2 (8.1)where we have neglected channel-length modulation. The drain current of Q1 is supplied byVDD through resistor R, which in most cases would be outside the IC chip. Since the gatecurrents are zero,I D1 = IREF = VDD − VGSR (8.2)where the current through R is considered to be the reference current of the current source andis denoted IREF . Equations (8.1) and (8.2) can be used to determine the value required for R.Now consider transistor Q2 : It has the same VGS as Q1 ; thus, if we assume that it is operatingin saturation, its drain current, which is the output current IO of the current source, will beIO = ID2 = 12 k′n( WL)2(V GS −Vtn)2 (8.3)where we have neglected channel-length modulation. Equations (8.1) and (8.3) enable us torelate the output current IO to the reference current IREF as follows:I OIREF= (W/L)2(W/L)1

Withdrawal of groundwater often causes subsidence, a situation whereMultiple Choiceaquifer volume decreases sharply.withdrawals exceed recharge rates.water quality and water tables fall.porous rocks settle and ground level falls.wells dry up.

The process of  reduction of threshold voltage of the transistor at higher drain voltages is called as